π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

The story of

πand related puzzles

Narrator: Niraj KhareCarnegie Mellon University Qatar

Being with math is being with the truth and eternity!

Nov, 15, 2017 1 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Time line III (a): series expressions for π

• Ludolph Van Ceulen using archimedean method with 500million sides calculated π calculated π to an accuracy of 20decimal digits by 1596. By the time he died in 1610, heaccurately found 35 digits! The digits were carved into histombstone.

2 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Time line III (a): series expressions for π

• Ludolph Van Ceulen using archimedean method with 500million sides calculated π calculated π to an accuracy of 20decimal digits by 1596. By the time he died in 1610, heaccurately found 35 digits! The digits were carved into histombstone.

2 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Ludolph Van Ceulen:3.14159265358979323846264338327950288...

Ludolph van CeulenDutch-German mathematicianLudolph van Ceulen was a German-Dutch mathematician fromHildesheim. He emigrated to the Netherlands. Wikipedia:

Born:January 28, 1540, Hildesheim, GermanyDied:December 31, 1610, Leiden, NetherlandsKnown for: piInstitution: Leiden UniversityNotable student: Willebrord Snellius

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000years ago!

• Around 450 BCE, Anaxagoras proposes ‘squaring thecircle’ from a prison! The puzzle was finally ‘settled’ in1882 AD.

• Around 250 BC, Archimedes proves that3.1408 < 310

71 = 22371 < π < 220

70 = 317 ≈ 3.1428.

4 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000years ago!

• Around 450 BCE, Anaxagoras proposes ‘squaring thecircle’ from a prison! The puzzle was finally ‘settled’ in1882 AD.

• Around 250 BC, Archimedes proves that3.1408 < 310

71 = 22371 < π < 220

70 = 317 ≈ 3.1428.

4 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000years ago!

• Around 450 BCE, Anaxagoras proposes ‘squaring thecircle’ from a prison! The puzzle was finally ‘settled’ in1882 AD.

• Around 250 BC, Archimedes proves that3.1408 < 310

71 = 22371 < π < 220

70 = 317 ≈ 3.1428.

4 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A rational number

An integer is a whole number (not a fractional number) thatcan be positive, negative, or zero.Examples:{· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }.

Definition

A rational number is a ratio of two integers. In other words, anumber q is a rational if there are integers a and b 6= 0 suchthat q = a

b .

Examples: 3.1 = 3110 , 3 = 3

1 , −5 = −51 , 3.127127127 · · · 127

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A rational number

An integer is a whole number (not a fractional number) thatcan be positive, negative, or zero.Examples:{· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }.

Definition

A rational number is a ratio of two integers.

In other words, anumber q is a rational if there are integers a and b 6= 0 suchthat q = a

b .

Examples: 3.1 = 3110 , 3 = 3

1 , −5 = −51 , 3.127127127 · · · 127

5 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A rational number

An integer is a whole number (not a fractional number) thatcan be positive, negative, or zero.Examples:{· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }.

Definition

A rational number is a ratio of two integers. In other words, anumber q is a rational if there are integers a and b 6= 0 suchthat q = a

b .

Examples: 3.1 = 3110 ,

3 = 31 , −5 = −5

1 , 3.127127127 · · · 127

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A rational number

An integer is a whole number (not a fractional number) thatcan be positive, negative, or zero.Examples:{· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }.

Definition

A rational number is a ratio of two integers. In other words, anumber q is a rational if there are integers a and b 6= 0 suchthat q = a

b .

Examples: 3.1 = 3110 , 3 = 3

1 ,

−5 = −51 , 3.127127127 · · · 127

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A rational number

An integer is a whole number (not a fractional number) thatcan be positive, negative, or zero.Examples:{· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }.

Definition

A rational number is a ratio of two integers. In other words, anumber q is a rational if there are integers a and b 6= 0 suchthat q = a

b .

Examples: 3.1 = 3110 , 3 = 3

1 , −5 = −51 ,

3.127127127 · · · 127

5 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A rational number

An integer is a whole number (not a fractional number) thatcan be positive, negative, or zero.Examples:{· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }.

Definition

A rational number is a ratio of two integers. In other words, anumber q is a rational if there are integers a and b 6= 0 suchthat q = a

b .

Examples: 3.1 = 3110 , 3 = 3

1 , −5 = −51 , 3.127127127 · · · 127

5 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A rational number

An integer is a whole number (not a fractional number) thatcan be positive, negative, or zero.Examples:{· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }.

Definition

A rational number is a ratio of two integers. In other words, anumber q is a rational if there are integers a and b 6= 0 suchthat q = a

b .

Examples: 3.1 = 3110 , 3 = 3

1 , −5 = −51 , 3.127127127 · · · 127

5 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

3.127127127 · · · 127 is rational!

Proof.

Let x = 3.127127127 · · · 127.

1000x = 3127.127127 · · · 127

x = 3.127127127 · · · 127

⇒ 1000x− x = 3127− 3

⇒ 999x = 3124

⇒ x =3124

999.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

3.127127127 · · · 127 is rational!

Proof.

Let x = 3.127127127 · · · 127.

1000x = 3127.127127 · · · 127

x = 3.127127127 · · · 127

⇒ 1000x− x = 3127− 3

⇒ 999x = 3124

⇒ x =3124

999.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

3.127127127 · · · 127 is rational!

Proof.

Let x = 3.127127127 · · · 127.

1000x = 3127.127127 · · · 127

x = 3.127127127 · · · 127

⇒ 1000x− x = 3127− 3

⇒ 999x = 3124

⇒ x =3124

999.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

3.127127127 · · · 127 is rational!

Proof.

Let x = 3.127127127 · · · 127.

1000x = 3127.127127 · · · 127

x = 3.127127127 · · · 127

⇒ 1000x− x = 3127− 3

⇒ 999x = 3124

⇒ x =3124

999.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Not everything is rational!

Hippasus of Metapontum (/hpss/; Greek: , Hppasos; fl. 3rdcentury BC), was a Pythagorean philosopher. He was the firstone to claim that there were irrational numbers!

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Not everything is rational!

Hippasus of Metapontum (/hpss/; Greek: , Hppasos; fl. 3rdcentury BC), was a Pythagorean philosopher. He was the firstone to claim that there were irrational numbers!

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Quotes

• The oldest, shortest words “yes” and “no” are thosewhich require the most thought. - Pythagoras

• A statement or a proposition in mathematics is a sentencethat is either true or false

but not both.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Quotes

• The oldest, shortest words “yes” and “no” are thosewhich require the most thought. - Pythagoras

• A statement or a proposition in mathematics is a sentencethat is either true or false but not both.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Quotes

• The oldest, shortest words “yes” and “no” are thosewhich require the most thought. - Pythagoras

• A statement or a proposition in mathematics is a sentencethat is either true or false but not both.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

An engineer or a manager?

Tom, Mary and Alice work for ‘Logic is Fun’. Two of them areengineers and exactly one of them is a manager. The manageralways lies and the engineers always speak the truth.

Who isthe manager?

Tom: “Mary is an engineer.”Alice: “Tom is the manager.”

Mary: “Alice is the manager.”Alice: “Mary is the manager.”

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

An engineer or a manager?

Tom, Mary and Alice work for ‘Logic is Fun’. Two of them areengineers and exactly one of them is a manager. The manageralways lies and the engineers always speak the truth. Who isthe manager?

Tom: “Mary is an engineer.”Alice: “Tom is the manager.”

Mary: “Alice is the manager.”Alice: “Mary is the manager.”

9 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

An engineer or a manager?

Tom, Mary and Alice work for ‘Logic is Fun’. Two of them areengineers and exactly one of them is a manager. The manageralways lies and the engineers always speak the truth. Who isthe manager?

Tom: “Mary is an engineer.”Alice: “Tom is the manager.”

Mary: “Alice is the manager.”Alice: “Mary is the manager.”

9 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

An engineer or a manager?

Tom, Mary and Alice work for ‘Logic is Fun’. Two of them areengineers and exactly one of them is a manager. The manageralways lies and the engineers always speak the truth. Who isthe manager?

Tom: “Mary is an engineer.”Alice: “Tom is the manager.”

Mary: “Alice is the manager.”Alice: “Mary is the manager.”

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A proof by contradiction!

To prove: Alice is the manager.

Proof.

On the contrary assume that Alice is not the manager.

Therefore, Alice is an engineer. Thus, she always speak thetruth. Hence,Tom and Mary are both managers. A contradiction!Therefore, our assumption must be false. So the negation of ourassumption is true.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A proof by contradiction!

To prove: Alice is the manager.

Proof.

On the contrary assume that Alice is not the manager.Therefore, Alice is an engineer. Thus, she always speak thetruth.

Hence,Tom and Mary are both managers. A contradiction!Therefore, our assumption must be false. So the negation of ourassumption is true.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A proof by contradiction!

To prove: Alice is the manager.

Proof.

On the contrary assume that Alice is not the manager.Therefore, Alice is an engineer. Thus, she always speak thetruth. Hence,Tom and Mary are both managers.

A contradiction!Therefore, our assumption must be false. So the negation of ourassumption is true.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A proof by contradiction!

To prove: Alice is the manager.

Proof.

On the contrary assume that Alice is not the manager.Therefore, Alice is an engineer. Thus, she always speak thetruth. Hence,Tom and Mary are both managers. A contradiction!

Therefore, our assumption must be false. So the negation of ourassumption is true.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A proof by contradiction!

To prove: Alice is the manager.

Proof.

On the contrary assume that Alice is not the manager.Therefore, Alice is an engineer. Thus, she always speak thetruth. Hence,Tom and Mary are both managers. A contradiction!Therefore, our assumption must be false. So the negation of ourassumption is true.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A proof by contradiction!

To prove: Alice is the manager.

Proof.

On the contrary assume that Alice is not the manager.Therefore, Alice is an engineer. Thus, she always speak thetruth. Hence,Tom and Mary are both managers. A contradiction!Therefore, our assumption must be false. So the negation of ourassumption is true.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Contrapositive

If president, then at least 35 years old.

If not yet 35, then cannot be the president.

If Tom is a parrot, then Tom is a bird.If Tom is not a bird then Tom cannot be a parrot.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Contrapositive

If president, then at least 35 years old.If not yet 35, then cannot be the president.

If Tom is a parrot, then Tom is a bird.If Tom is not a bird then Tom cannot be a parrot.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Contrapositive

If president, then at least 35 years old.If not yet 35, then cannot be the president.

If Tom is a parrot, then Tom is a bird.If Tom is not a bird then Tom cannot be a parrot.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Contrapositive

If president, then at least 35 years old.If not yet 35, then cannot be the president.

If Tom is a parrot,

then Tom is a bird.If Tom is not a bird then Tom cannot be a parrot.

11 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Contrapositive

If president, then at least 35 years old.If not yet 35, then cannot be the president.

If Tom is a parrot, then Tom is a bird.

If Tom is not a bird then Tom cannot be a parrot.

11 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Contrapositive

If president, then at least 35 years old.If not yet 35, then cannot be the president.

If Tom is a parrot, then Tom is a bird.If Tom is not a bird then Tom cannot be a parrot.

11 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Contrapositive

If president, then at least 35 years old.If not yet 35, then cannot be the president.

If Tom is a parrot, then Tom is a bird.If Tom is not a bird then Tom cannot be a parrot.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

History of π is irrational.

• There are conflicting claims who first ‘guessed’ that π is nota rational.But was believed by many by 5-th century AD.

• In 1761 to 1776, Lambert and Legendre proved that π isnot a ratio of two integers.[Cajori, page 246]

• In 1882, Ferdinand von Lindemann proved transcendenceof π (i.e., squaring the circle is impossible). [Berggren, page407]

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

History of π is irrational.

• There are conflicting claims who first ‘guessed’ that π is nota rational.But was believed by many by 5-th century AD.

• In 1761 to 1776, Lambert and Legendre proved that π isnot a ratio of two integers.[Cajori, page 246]

• In 1882, Ferdinand von Lindemann proved transcendenceof π (i.e., squaring the circle is impossible). [Berggren, page407]

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

History of π is irrational.

• There are conflicting claims who first ‘guessed’ that π is nota rational.But was believed by many by 5-th century AD.

• In 1761 to 1776, Lambert and Legendre proved that π isnot a ratio of two integers.[Cajori, page 246]

• In 1882, Ferdinand von Lindemann proved transcendenceof π (i.e., squaring the circle is impossible). [Berggren, page407]

12 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

History of π is irrational.

• There are conflicting claims who first ‘guessed’ that π is nota rational.But was believed by many by 5-th century AD.

• In 1761 to 1776, Lambert and Legendre proved that π isnot a ratio of two integers.[Cajori, page 246]

• In 1882, Ferdinand von Lindemann proved transcendenceof π (i.e., squaring the circle is impossible). [Berggren, page407]

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Lambart’s idea: I

Lemma

Let x be a real number. If x is rational, then tan(x) isirrational.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Lambart’s idea: II

Theorem

π is irrational.

Proof.

Assume that π = ab where a and b are integers. Thus,

π4 = a

4b .

By previous lemma, tan(π4 ) must be an irrational. Buttan(π4 ) = 1 = 1

1 . A contradiction!

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Lambart’s idea: II

Theorem

π is irrational.

Proof.

Assume that π = ab where a and b are integers. Thus,

π4 = a

4b .By previous lemma, tan(π4 ) must be an irrational.

Buttan(π4 ) = 1 = 1

1 . A contradiction!

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Lambart’s idea: II

Theorem

π is irrational.

Proof.

Assume that π = ab where a and b are integers. Thus,

π4 = a

4b .By previous lemma, tan(π4 ) must be an irrational. Buttan(π4 ) = 1 = 1

1 . A contradiction!

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Lambart’s idea: II

Theorem

π is irrational.

Proof.

Assume that π = ab where a and b are integers. Thus,

π4 = a

4b .By previous lemma, tan(π4 ) must be an irrational. Buttan(π4 ) = 1 = 1

1 . A contradiction!

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Theorem

π is irrational.

‘Irrationality is not limited to numbers!’

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

On the contrary assume that π is a rational number.Thus, π = a

b for integers a and b 6= 0. Without loss of generality,let a and b be both positive.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

On the contrary assume that π is a rational number.Thus, π = a

b for integers a and b 6= 0. Without loss of generality,let a and b be both positive.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

On the contrary assume that π is a rational number.Thus, π = a

b for integers a and b 6= 0. Without loss of generality,let a and b be both positive.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

On the contrary assume that π is a rational number.Thus, π = a

b for integers a and b 6= 0. Without loss of generality,let a and b be both positive.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A bunny found!

For any positive integer n, define

f(x) =bnxn(π − x)n

n!=xn(a− bx)n

n!

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A bunny found!

For any positive integer n, define

f(x) =bnxn(π − x)n

n!=xn(a− bx)n

n!

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

A bunny found!

For any positive integer n, define

f(x) =bnxn(π − x)n

n!=xn(a− bx)n

n!

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Some properties of f(x): I

• For all real x, f(x) = f(π − x).

• For any non-negative integer k, f (k)(π − x) = −1kf (k)(x)where f (k)(x) denotes the k-th derivative of f with respectto x.

• For all non-negative integer k, f (k)(0) is an integer.Therefore, for all non-negative integer k, f (k)(π) is aninteger too.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Some properties of f(x): I

• For all real x, f(x) = f(π − x).

• For any non-negative integer k, f (k)(π − x) = −1kf (k)(x)where f (k)(x) denotes the k-th derivative of f with respectto x.

• For all non-negative integer k, f (k)(0) is an integer.

Therefore, for all non-negative integer k, f (k)(π) is aninteger too.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Some properties of f(x): I

• For all real x, f(x) = f(π − x).

• For any non-negative integer k, f (k)(π − x) = −1kf (k)(x)where f (k)(x) denotes the k-th derivative of f with respectto x.

• For all non-negative integer k, f (k)(0) is an integer.Therefore, for all non-negative integer k, f (k)(π) is aninteger too.

18 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Some properties of f(x): I

• For all real x, f(x) = f(π − x).

• For any non-negative integer k, f (k)(π − x) = −1kf (k)(x)where f (k)(x) denotes the k-th derivative of f with respectto x.

• For all non-negative integer k, f (k)(0) is an integer.Therefore, for all non-negative integer k, f (k)(π) is aninteger too.

18 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Some properties of f(x): I

• For all real x, f(x) = f(π − x).

• For any non-negative integer k, f (k)(π − x) = −1kf (k)(x)where f (k)(x) denotes the k-th derivative of f with respectto x.

• For all non-negative integer k, f (k)(0) is an integer.Therefore, for all non-negative integer k, f (k)(π) is aninteger too.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

f (k)(0) is an integer: I

f(x) =xn(a− bx)n

n!

=xn

n!

n∑i=0

((n

i

)a(n−i)(−1)ibixi

)

=

n∑i=0

((ni

)a(n−i)(−1)ibixn+i

)n!

As f(x) is a polynomial of degree 2n, for all k > 2n f (k)(x) = 0.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

f (k)(0) is an integer: II

f (k)(x)

=

n∑i=0

c a(n−i)(−1)ibi(n+ i)(n+ i− 1) · · · (n+ i− {k − 1})xn+i−k

n!

where c =(ni

)and when n+ i ≥ k.

When x = 0, only the term with n+ i = k contributes. Inparticular f (k)(0) = 0 for all k < n. For k = n+ i,

f (k)(0)

=c a(n−i)(−1)ibi(k)(k − 1) · · · (k − {k − 1})

n!

=c a(n−i)(−1)ibi(n+ i)(n+ i− 1) · · · (n) · · · (1)

n!

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Some properties of f(x): II

Define g(x) = f(x)− f (2)(x) + f (4)(x)− f (6)(x) + · · ·+(−1)kf (2k)(x) + · · ·+ (−1)(n−1)f (2(n−1))(x) + (−1)nf (2n)(x).Thus, g(0) and g(π) are integers.

Note that g(x) + g(2)(x) = f(x) where g(2)(x) = g′′(x).

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Some properties of f(x): II

Define g(x) = f(x)− f (2)(x) + f (4)(x)− f (6)(x) + · · ·+(−1)kf (2k)(x) + · · ·+ (−1)(n−1)f (2(n−1))(x) + (−1)nf (2n)(x).Thus, g(0) and g(π) are integers.Note that g(x) + g(2)(x) = f(x) where g(2)(x) = g′′(x).

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Some properties of f(x): II

Define g(x) = f(x)− f (2)(x) + f (4)(x)− f (6)(x) + · · ·+(−1)kf (2k)(x) + · · ·+ (−1)(n−1)f (2(n−1))(x) + (−1)nf (2n)(x).Thus, g(0) and g(π) are integers.Note that g(x) + g(2)(x) = f(x) where g(2)(x) = g′′(x).

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Some properties of f(x): II

Define g(x) = f(x)− f (2)(x) + f (4)(x)− f (6)(x) + · · ·+(−1)kf (2k)(x) + · · ·+ (−1)(n−1)f (2(n−1))(x) + (−1)nf (2n)(x).Thus, g(0) and g(π) are integers.Note that g(x) + g(2)(x) = f(x) where g(2)(x) = g′′(x).

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Anti-derivative of f(x)sin(x)

d

dx{g′(x)sin(x)− g(x)cos(x))}

= g′′(x)sin(x) + g′(x)cos(x)− g′(x)cos(x) + g(x)sin(x)

= g′′(x)sin(x) + g(x)sin(x)

= {g′′(x) + g(x)}sin(x)

= f(x)sin(x)

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is an integer for all positive integers n.

∫ π

0f(x)sin(x)dx

= {g′(x)sin(x)− g(x)cos(x)}|π0= {g′(π)sin(π)− g(π)cos(π)} − {g′(0)sin(0)− g(0)cos(0)}

= g(π) + g(0)

Hence,∫ π0 f(x)sin(x)dx is an integer for all positive integers n.

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π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is an integer for all positive integers n.

∫ π

0f(x)sin(x)dx

= {g′(x)sin(x)− g(x)cos(x)}|π0= {g′(π)sin(π)− g(π)cos(π)} − {g′(0)sin(0)− g(0)cos(0)}= g(π) + g(0)

Hence,∫ π0 f(x)sin(x)dx is an integer for all positive integers n.

23 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is an integer for all positive integers n.

∫ π

0f(x)sin(x)dx

= {g′(x)sin(x)− g(x)cos(x)}|π0= {g′(π)sin(π)− g(π)cos(π)} − {g′(0)sin(0)− g(0)cos(0)}= g(π) + g(0)

Hence,∫ π0 f(x)sin(x)dx is an integer for all positive integers n.

23 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is an integer for all positive integers n.

∫ π

0f(x)sin(x)dx

= {g′(x)sin(x)− g(x)cos(x)}|π0= {g′(π)sin(π)− g(π)cos(π)} − {g′(0)sin(0)− g(0)cos(0)}= g(π) + g(0)

Hence,∫ π0 f(x)sin(x)dx is an integer for all positive integers n.

23 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is not an integer for LARGE n.

∫ π

0f(x)sin(x)dx

=

∫ π

0

bnxn(π − x)n

n!sin(x)dx

=bn

n!

∫ π

0xn(π − x)nsin(x)dx

24 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is not an integer for LARGE n.

0 <bn

n!

∫ π

0xn(π − x)nsin(x)dx

25 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is not an integer for LARGE n.

0 <bn

n!

∫ π

0xn(π − x)nsin(x)dx

25 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

War between exponent and factorial!

0 <bn

n!

(π2

)2n(π)

y =6n

n!

26 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

War between exponent and factorial!

0 <bn

n!

(π2

)2n(π)

y =6n

n!

26 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

War between exponent and factorial!

0 <bn

n!

(π2

)2n(π)

y =6n

n!

26 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is not an integer for LARGE n: III

As n→∞,

limn→∞

bn

n!

(π2

)2n(π) = 0.

0 <bn

n!

∫ π

0xn(π − x)nsin(x)dx

<bn

n!

(π2

)2n(π)

< 1.

27 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is not an integer for LARGE n: III

As n→∞,

limn→∞

bn

n!

(π2

)2n(π) = 0.

0 <bn

n!

∫ π

0xn(π − x)nsin(x)dx

<bn

n!

(π2

)2n(π)

< 1.

27 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is not an integer for LARGE n: III

As n→∞,

limn→∞

bn

n!

(π2

)2n(π) = 0.

0 <bn

n!

∫ π

0xn(π − x)nsin(x)dx

<bn

n!

(π2

)2n(π)

< 1.

27 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is not an integer for LARGE n: III

As n→∞,

limn→∞

bn

n!

(π2

)2n(π) = 0.

0 <bn

n!

∫ π

0xn(π − x)nsin(x)dx

<bn

n!

(π2

)2n(π)

< 1.

27 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof∫ π

0 f(x)sin(x)dx is not an integer for LARGE n: III

As n→∞,

limn→∞

bn

n!

(π2

)2n(π) = 0.

0 <bn

n!

∫ π

0xn(π − x)nsin(x)dx

<bn

n!

(π2

)2n(π)

< 1.

27 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Bunny or a pigeon?

For large integer n,

• bn

n!

∫ π0 x

n(π − x)nsin(x)dx is an integer.

•0 <

bn

n!

∫ π

0xn(π − x)nsin(x)dx < 1.

28 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Formal and informal references:Informal References

• Documentaries:• Math and rise of civilizations:http://motionpic.com/catalogue/

math-rise-of-civilization-science-docs-documentaries-2/• BBC: Story of Mathematics: http://www.bbc.co.uk/programmes/b00dxjls/episodes/guide

• Websites:• The history of pi by David Wilsonhttp://sites.math.rutgers.edu/~cherlin/History/

Papers2000/wilson.html• Archimedes’ Approximation of Pihttp://itech.fgcu.edu/faculty/clindsey/mhf4404/

archimedes/archimedes.html• Euclid’s Elementshttps://mathcs.clarku.edu/~djoyce/java/elements/

• Wekipedia: Ludolph Van Ceulen’s biographyhttps://en.wikipedia.org/wiki/Ludolph_van_Ceulen 29 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Formal and informal references:Formal References(I)

George E. Andrews, Peter Paule. Some questions concerningcomputer-generated proofs of a binomial double-sumidentity, J. Symbolic Comput. 16(1993), 147–151.

P. Backman, The history of Pi, The Golem Press. BoulderColorado, 1971.

J.L.Berggren, J. , Borwein, P. Borwein, Pi: A Source Book,Springer, 2004.

D. Blatner, The joy of Pi, Walker Publishing Company, IncNewyork, 1997.

F. Cajori, A history of Mathematics, MacMillan and Co.London, 1926.

Sir T. Heath,A History of Greek Mathematics: From Thalesto Euclid, Volume 1, Dover Publications, inc. Newyork,1981.

J. J. O’Connor, E.F. Robertson, The MacTutor History ofMathematics Archive,World Wide Web., 1996.

30 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Formal and informal references:Formal References (II)

Bourbaki, N Fonctions d’une variable relle, chap. IIIIII,Actualits Scientifiques et Industrielles (in French), 1074,Hermann, pp. 137138, 1949.

Jeffreys, Harold, Scientific Inference (3rd ed.), CambridgeUniversity Press, p. 268, 1973.

Niven, Ivan, ”A simple proof that is irrational” (PDF),Bulletin of the American Mathematical Society, 53 (6), p.509, 1947.

31 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Acknowledgments

I would specially like to thank:i) Prof. Marion Oliver for his support, interest in history ofmathematics and encouragement for the concept of ‘ExploreMath’.ii) Prof. H. Demirkoparan and Prof. Z. Yilma for their help andsuggestions.iii) Kara, Angela, Catalina and Geetha for promoting the eventand taking care of logistics.iv) Ghost of Ludolph Van Ceulen for haunting me and pushingme to explore more about Pi :).

32 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Something to carry home!

Q: What will a logician choose: a half of an egg or eternal bliss?

A: A half of an egg! Because nothing is better than eternalbliss, and a half of an egg is better than nothing.

33 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Something to carry home!

Q: What will a logician choose: a half of an egg or eternal bliss?

A: A half of an egg! Because nothing is better than eternalbliss, and a half of an egg is better than nothing.

33 / 33

π is irrationalPhilosophy of MathematicsNiven and Bourbaki’s proof

Something to carry home!

Q: What will a logician choose: a half of an egg or eternal bliss?

A: A half of an egg! Because nothing is better than eternalbliss, and a half of an egg is better than nothing.

33 / 33